ECE35_SP17_HW03_Solution.pdf

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ECE 35

Homework #3 Solution (Spring 2015, Taur)

All homework problems come from the textbook, “Introduction to Electric Circuits”, by

Svoboda & Dorf, 9

th

Edition. Question number in the 8

th

edition is listed for reference.

Question Number

Svoboda & Dorf, 8th Edition

Svoboda & Dorf, 9th Edition

1

P 4.3-3

P 4.3-3

2

P 4.3-7

P 4.3-7

3

P 4.3-9

P 4.3-9

4

P 4.3-12

P 4.3-12

5

P 4.4-5

P 4.4-5

6

P 4.4-11

P 4.4-11

7

P 4.4-15

P 4.4-14

8

P 4.5-2

P 4.5-2

9

P 4.5-6

P 4.5-6

10

P 4.6-5

P 4.6-5

11

P 4.6-8

P 4.6-8

12

P 4.7-7

P 4.7-7

13

P 4.7-13

P 4.7-12

14

P 4.11-2

P 4.11-2

(2)

Figure P4.3-3

P4.3-3. Determine the values of the power supplied by each of the sources in the circuit shown in Figure P4.3-3.

Solution: First, label the node voltages, and express the resistor currents in terms of the node voltages.

Identify the super node corresponding to the 24 V source

Apply KCL to the super node to get

a

a a

a a

12

24

24

0.6

196

6

32 V

10

40

40

v

v

v

v

v

The 12 V source supplies

12

12

a

24

12

12

32 24

4.8 W

10

10

v

 

The 24 V source supplies

a

32

24

0.6

24

0.6

4.8 W

40

40

v

(3)

P 4.3-7

Determine the values of the node voltages, v1

and

v2, in Figure P 4.3-7. Determine the values of the

currents ia and ib.

Figure P 4.3-7

Solution:

Apply KCL at nodes 1 and 2 to get

1 1 1 2

1 2

10

23 3 150 1000 3000 5000

v v v v

v v

 

    

2 1 3 3

1 3

10

-4 19 50 4000 5000 2000

v v v v

v v

 

    

Solving, e.g. using MATLAB, gives

1

1 1

2

23

3

150

7.06 V and

4.12 V

4 19

50

v

v

v

v

 

 

 

Then b 1 2

7.06 4.12

0.588 mA

5000 5000

v v

i     

Apply KCL at the top node to get

1 2

a

10 10 7.06 10 4.12 10

4.41 mA

1000 4000 1000 4000

v v

(4)

P 4.3-9

Determine the values of the node voltages of the circuit shown in Figure P 4.3-9.

Figure P 4.3-9

Solution:

Express the voltage source voltages as functions of the node voltages to get

2 1 5 and 4 15

vvv

Apply KCL to the supernode corresponding to the 5 V source to get

1 3 2

1 2 3

15

1.25 0 80 5 2 5

8 20

v v v

v v v

 

      

Apply KCL at node 3 to get

1 3 3 3

1 3

15

15 28 150

8 40 12

v v v v

v v

 

     

Solving, e.g. using MATLAB, gives

1 1

2 2

3 3

1 1 0 5 22.4

5 2 5 80 27.4

15 0 28 150 17.4

v v

v v

v v

   

     

   

   

   

     

   

     

        

So the node voltages are:

1 22.4 V, 2 27.4 V, 3 17.4 V, and 4 15

(5)

P 4.3-12

Determine the values of the node voltages of the circuit shown in Figure P 4.3-12.

Figure P 4.3-12

Solution:

Express the voltage source voltages in terms of the node voltages:

2 1 8 and 3 1 12

vvvv

Apply KVL to the supernode to get

2 1 3

2 1 3

0 2 5 4 0

10 4 5

v v v

v v v

      

so

1

1

1

1

64

2 8 5 4 12 0 V

11

v v v v

       

The node voltages are

1

2

3

5.818 V

2.182 V

6.182 V

v v v

(6)

P 4.4-5

Determine the value of the current

ix in the

circuit of Figure P 4.4-5.

Answer: i

x = 2.4 A

Figure P 4.4-5

Solution:

First, express the controlling current of the CCVS in terms of the node voltages: 2

2

x

v i

Next, express the controlled voltage in terms of the node voltages:

2

2 2

24

12 3 3 V

2 5

x

v

v i v

    

(7)

P 4.4-11

Determine the power supplied by the

dependent source in the circuit shown in Figure P

4.4-11.

Figure P 4.4-11

Solution:

This circuit contains two ungrounded voltage sources, both incident to node x. In such a circuit it is necessary to merge the supernodes corresponding to the two ungrounded voltage sources into a single supernode. That single supernode separates the two voltage sources and their nodes from the rest of the circuit. It consists of the two resistors and the current source. Apply KCL to this supernode to get

x x

x

20

4 0 10 V

2 10

v v

v

    

.

The power supplied by the dependent source is

0.1vx

30

 30 W

.

P 4.4-14

The voltages

v1, v2,

v3, and v4 are

the node voltages corresponding to nodes 1, 2,

3, and 4 in Figure P 4.4-14. Determine the

values of these node voltages.

(8)

Solution:

Express the controlling voltage and current of the dependent sources in terms of the node voltages:

a 4

vv and b 3 4 2

v

v

i

R

Express the voltage source voltages in terms of the node voltages:

1 s

vV and v2v3 Ava Av4

Apply KCL to the supernode corresponding to the dependent voltage source

2 1 3 4

s 2 1 2 2 1 3 1 4 1 2 s

1 2

v

v

v

v

I

R v

R v

R v

R v

R R I

R

R

 

Apply KCL at node 4:

3 4 3 4 4 2

3 4

2 2 3 3

1 1 0

v v v v v R

B B v B v

R R R R

 

 

       

 

Organizing these equations into matrix form:

1 s

2

2 2 1 1

3 1 2 s

2 4 3

1 0 0 0

0 1 1

0

0 0 1 1 0

v V

A

v

R R R R

v R R I

R v B B R                                                

With the given values:

1 1

2 2

3 3

4 4

1

0

0

0

25

25

0

1

1

5

0

44.4

20

20 10

10

400

8.4

0

0

4

4.667

0

7.2

(9)

P 4.5-2

The values of the mesh currents in the

circuit shown in Figure P 4.5-2 are

i1 = 2 A, i2 = 3 A, and i3 = 4 A.

Determine the values of the resistance R and of

the voltages v1 and v2 of the voltage sources.

Answers: R

= 12 Ω, v1 = –4 V, and v2 = –28 V

Figure P 4.5-2

Solution:

The mesh equations are:

Top mesh: 4 (2 3) R(2) 10 (2 4)0 so R = 12 .

Bottom, right mesh: 8 (4 3) 10 (4 2)   v2 0

so v2 = 28 V.

Bottom left mesh  v1 4 (3 2) 8 (3 4)   0

(10)

P 4.5-6

Simplify the circuit shown in

Figure P 4.5-6 by replacing series and

parallel resistors by equivalent resistors.

Next, analyze the simplified circuit by

writing and solving mesh equations.

(a)

Determine the power supplied by

each source.

(b)

Determine the power absorbed by the

30-Ω resistor.

Figure P 4.5-6

Solution:

Replace series and parallel resistors with equivalent resistors:

60  300  50 

,

40



60

 

100

and

100

 

30

 

80

560

 

200

so the simplified circuit is

The mesh equations are

1 1 2

200i 50 ii 120

2 1 2

100i  8 50 ii 0 or

1 1

2 2

250

50

12

0.04

50

150

8

0.04

i

i

i

i

 

 

 

 

 

 

 

 

 

The power supplied by the 12 V source is

12

i

1

12 0.04

0.48 W

. The power supplied by the 8 V

source is

8

i

2

  

8

0.04

0.32 W

. The power absorbed by the 30  resistor is

  

  

2 2

1 30 0.04 30 0.048 W

(11)

P 4.6-5

Determine the value of the voltage measured by the voltmeter in Figure P 4.6-5.

Answer:

8 V

Figure P 4.6-5

Solution:

Label the mesh currents:

Express the current source current in terms of the mesh currents:

3 1 2 1 3 2

i  ii  i

Supermesh: 6i13i35

i2i3

 8 0  6i15i28i38

Lower, left mesh:   12 8 5

i2i3

0  5i2  4 5i3

Eliminating i1 and i2 from the supermesh equation:

3

 

3

3 3

6 i   2 4 5i 8i 8  9i 24

The voltage measured by the meter is: 3

24

3 3 8 V

9

i  

(12)

P 4.6-8

Determine values of the mesh

currents, i1, i2, and i3, in the circuit shown in

Figure P 4.6-8.

Figure P 4.6-8

Solution:

Use units of V, mA and k

. Express the currents to the supermesh to get

1 3 2

i  i

Apply KVL to the supermesh to get

3 3

 

3

 

1 2

1 2 3

4 ii  1 i  3 1 ii 0  i 5i 5i 3

Apply KVL to mesh 2 to get

 

 

2 2 3 2 1 1 2 3

2i 4 ii  1 ii 0  1 i 7i 4i 0

Solving, e.g. using MATLAB, gives

1 1

2 2

3 3

1 0 1 2 3

1 5 5 3 1

1 7 4 0 1

i i

i i

i i

   

     

   

   

   

     

   

      

(13)

P 4.7-7

The currents

i1,

i2 and

i3 are the

mesh currents of the circuit shown in Figure

P 4.7-7. Determine the values of i1, i2, and i3.

Figure P 4.7-7

Solution:

Express va and ib, the controlling voltage and current of the dependent sources, in terms of the mesh

currents

a 5 2 3 and b 2

vii i  i

Next express 20 ib and 3 va, the controlled voltages of the dependent sources, in terms of the mesh

currents

b 2

20 i  20 i

and

3 va 15

i2i3

Apply KVL to the meshes

2 3

 

2

1

15 i i 20 i 10 i 0

     

20 i2

 

5 i2 i3

20 i2 0

     

2 3

2 3

10 5 ii 15 ii 0

These equations can be written in matrix form

1

2

3

10 35 15 0

0 45 5 0

0 10 10 10

i i i

  

   

 

 

 

   

 

    

    

Solving, e.g. using MATLAB, gives

1 1.25 A, 2 0.125 A, and 3 1.125 A

(14)

P 4.7-12

The currents i1, i2, and i3 are the

mesh currents corresponding to meshes 1,

2, and 3 in Figure P 4.7-12. Determine the

values of these mesh currents.

Figure P 4.7-12

Solution:

Express the controlling voltage and current of the dependent sources in terms of the mesh currents:

a 3 1 2

vR ii and ib  i3 i2

Express the current source currents in terms of the mesh currents:

2 s

i  I and i1 i3 B ib B i

3i2

Consequently

1

1

3 s

i

B

i

B I

Apply KVL to the supermesh corresponding to the dependent current source

R i1 3A R i3

1i2

R2

i3i2

R i3

1i2

Vs 0

OR

A1

R i3 1

R2

A1

R i3

 

2R1R2

i3Vs

Organizing these equations into matrix form:

1 s

2 s

3 s

3 2 3 1 2

0 1 0

1 0 1

1 1

i I

B i B I

i V

A R R A R R R

                               

With the given values:

1 1

2 2

3 3

0 1 0 2 0.8276

1 0 4 6 2 A

60 80 50 25 1.7069

(15)

P 4.11-2

An old lab report asserts that the

node voltages of the circuit of Figure P

4.10-2 are

va = 4 V, vb = 20 V, and vc = 12 V.

Are these correct?

Solution:

Apply KCL at node a:

2 0

4 2

20 4 4

2 4 0

4 2

b a a

vv v

 

  

 

 

    

 

Figure

Figure P 4.3-7  Solution:

Figure P

4.3-7 Solution: p.3
Figure P 4.3-12  Solution:

Figure P

4.3-12 Solution: p.5
Figure P 4.7-7  Solution:

Figure P

4.7-7 Solution: p.13

References

Related subjects : two-to-one merge